"Greg Neill" escreveu na mensagem
...
"El Enrrabadore-mor" wrote in message


Wait, you're confused about the equivalence
principle and you confuse gravity with centrifugal
force?
I've a secret for you, gravity and centrifugal force
are not equivalent in nature. Gravity follows an
inverse SQUARED radius function. Centrifugal
follows an inverse LINEAR radius.

No! The inverse square relationship for gravity
is the consequence of a particular geometry of
the source mass. The field from a point source
or spherically symmetric source diverges. For
spherically symmetric sources the resulting field
varies as the inverse square.

Consider instead an infinite sheet of mass. In
such a case the field lines are parallel and the
potential drops linearly with distance. Or
consider the hypothetical hole bored through a
hypothetically uniform density planet (go pole
to pole to avoid coriolis complications). Thus,
as Einstein said, acceleration (centrifugal
included) is equivalent to a *uniform*
gravitational field.

Well, I'm not going to say you're wrong, nor right.
A discussion on the nature of gravity and inertia
will be always inconclusive, since nobody knows
the nature of none of them.

Yes, you're right about gravity. Gravity effects are a
consequence of a particular geometry of the source
mass. Gravity force is the integral of the contribution
of all infinitesimal masses. That's why gravity on
the center of the Earth is zero. That's why a mass
inside a spherical potential that goes like gravity by
an inverse square Law causes zero force and no
motion of the inside body. (That's Gauss Theorem).

For an infinite sheet of mass of zero thickness the
integral of the contribution of all infinitesimal masses
causes a parallel line field and the potential will drop
linearly with distance (due to the integral only).
For a solid sphere of mass it will be usual inverse
square Law.

Now, centrifugal force is due to inertia.
Inertia, like gravity, is something that one cannot
explain in terms of the cause.
Centrifugal force doesn't care about the mass
distribution. It can be a sphere or a sheet of mass.
All that matters is the Center of Mass and its distance
to the center of rotation.

Centrifugal force = m r w^2 = m V^2 / r


Only for
stable orbits, where a body can be considered
to be in free-fall (no forces), one can make
such approximation.

Any force that results in a uniform acceleration
will be equivalent to a uniform gravitational field.

Yes. "Equivalent" is all that you can say.
Gravity and inertia are not alike in nature, they just
become equivalent for symmetric distributions of
mass. Lucky that all masses around the Universe
are spherical and symmetric at large scales.


A time varying force would be equivalent to a time
varying gravitational field, which we don't find a
lot of in our day to day experience, but we can
certainly concoct hypothetical situations where it
can happen.

Yes, but I cannot see the point.

"El Enrrabadore-mor" wrote in message

"Greg Neill" escreveu na mensagem
...

Any force that results in a uniform acceleration
will be equivalent to a uniform gravitational field.

Yes. "Equivalent" is all that you can say.
Gravity and inertia are not alike in nature, they just
become equivalent for symmetric distributions of
mass. Lucky that all masses around the Universe
are spherical and symmetric at large scales.

The equivalence principle says that inertial mass
and gravitational mass are equal in magnitude for any
body with mass. It also imples that acceleration is
equivalent to gravitation for suitably chosen
gravitation and acceleration fields, the upshot being
that all effects caused by one should also be caused
by the other, such as time dilation in a potential
field.

A time varying force would be equivalent to a time
varying gravitational field, which we don't find a
lot of in our day to day experience, but we can
certainly concoct hypothetical situations where it
can happen.

Yes, but I cannot see the point.

You said that acceleration and gravity are not alike in
nature (because gravity follows an inverse square law).
I am pointing out that this is not necessarily true as
it is dependent on your choice of geometry.

Hi Elvis

On May 11, 4:38 am, "El Enrrabadore-mor"
wrote:
"Ken S. Tucker" escreveu na ...

Elvis, I think you're being unfair.
...
The right analysis is that time dilatation is a consequence
of the speed of light and the DISTANCE travelled by light.
That is, time dilatation only is required for observations made
at distance, by means of light, like Einstein said.
Not the actual bull**** that follows because physicists only
have a stone hammer to work with.

Hi, Steve Wonder.

Nope, I'm Rachel Welch in drag.

Our discussion has evolved to the transition
from SR to GR, in the weak field limit, that
was fairly much agreed upon in the early 20th
century, that have empirical support.

If the discussion has evolved to GR I didn't notice.
Where is it?

In the ref'd Pauli essay.

Wait, you're confused about the equivalence
principle and you confuse gravity with centrifugal
force?

Not really.

I've a secret for you, gravity and centrifugal force
are not equivalent in nature. Gravity follows an
inverse SQUARED radius function. Centrifugal
follows an inverse LINEAR radius. Only for
stable orbits, where a body can be considered
to be in free-fall (no forces), one can make
such approximation.

As a little brat I've done the experiment
personally, I suspended a bucket and felt
the force. Then I removed some stones from
the bucket and twirled it.
I exchanged g-force for c-force+g-force.
Just do the experiment and get the feel.

Most of the fella's have given more than an
adequate airing of your misunderstandings, in
fact I'm surprised you haven't been flamed to
a crisp, so why throw insults?
Ken
BTW, before lecturing us, learn to ****ing
spell "dilation", that's the 2nd ****ing
time I've told you that!

"Dilatation" follows natural from my native language,
and I can't well myself. Besides that the spell checker
is happy with that word, so... (thanks).

You're right! I checked out the dictionary
and "dilitation" is in there :-), but it's
never used that way in physics.
Ken

"Greg Neill" escreveu na mensagem
m...
"El Enrrabadore-mor" wrote in message

"Greg Neill" escreveu na mensagem
...

Any force that results in a uniform acceleration
will be equivalent to a uniform gravitational field.

Yes. "Equivalent" is all that you can say.
Gravity and inertia are not alike in nature, they just
become equivalent for symmetric distributions of
mass. Lucky that all masses around the Universe
are spherical and symmetric at large scales.

The equivalence principle says that inertial mass
and gravitational mass are equal in magnitude for any
body with mass. It also imples that acceleration is
equivalent to gravitation for suitably chosen
gravitation and acceleration fields, the upshot being
that all effects caused by one should also be caused
by the other, such as time dilation in a potential
field.

Now we agree.
The only true statement are that masses are the same
(in magnitude) for the Equivalence Principle.
Accelerations are not (a "true in nature" statement).

Therefore, like you've said, only "for suitably chosen
gravitation and acceleration fields accelerations are
equivalent."
Quoting: "It also imples that acceleration is
equivalent to gravitation for suitably chosen
gravitation and acceleration fields, ..."
End quote.
(The situation where the fields are equivalent
in acceleration we call them "orbits".)

Greg Neill, we can only agree, even if it looks
the contrary. The difference is that I'm free to
say it, you don't look so (but I can be wrong
of course).


A time varying force would be equivalent to a time
varying gravitational field, which we don't find a
lot of in our day to day experience, but we can
certainly concoct hypothetical situations where it
can happen.

Yes, but I cannot see the point.

You said that acceleration and gravity are not alike in
nature (because gravity follows an inverse square law).
I am pointing out that this is not necessarily true as
it is dependent on your choice of geometry.

If you understood that I'm sorry, for sure I didn't.

The "inverse square Law on radius " versus "inverse
linear Law" issue, it was about gravity versus centrifugal
force. That's and old story between me and Ken where
we've agree upon (the speed of gravity infinite for orbits
to be stable).

"Tom Roberts" escreveu na mensagem
...
El Enrrabadore-mor wrote:
"Tom Roberts" escreveu na mensagem
...
Look at the subject of this thread. "simplifying" to circular motion is
what is being discussed.

Such simplification makes your solution ridiculous, that's all.
It's not "ridiculous", it is what this thread is about. I gave a general
equation (in response to an inquiry), and then applied it to the subject
of this thread. shrug

If you find the thread subject (the original inquiry) ridiculous,

YOU are the one who claimed it was "ridiculous", not I. If this is the
level of your accuracy of reading and remembering what YOU wrote, then
discussing anything with you is hopeless....


I get the picture, thanks and sorry too.

"El Enrrabadore-mor" wrote in message


Therefore, like you've said, only "for suitably chosen
gravitation and acceleration fields accelerations are
equivalent."

By this I meant that for a given gravitational
field you choose an acceleration (applied by whatever
means) accordingly, and vice versa.

Quoting: "It also imples that acceleration is
equivalent to gravitation for suitably chosen
gravitation and acceleration fields, ..."
End quote.
(The situation where the fields are equivalent
in acceleration we call them "orbits".)

Orbit is not necessary. Take a given test mass and
place it in a gravitational field and measure the
resulting acceleration. Take the same test mass and
take it far away from sources of gravity and apply
a force that produces the same acceleration as
before. The resulting effects will be
indistinguishable.

"Greg Neill" escreveu na mensagem
m...
"El Enrrabadore-mor" wrote in message


Therefore, like you've said, only "for suitably chosen
gravitation and acceleration fields accelerations are
equivalent."

By this I meant that for a given gravitational
field you choose an acceleration (applied by whatever
means) accordingly, and vice versa.

Quoting: "It also imples that acceleration is
equivalent to gravitation for suitably chosen
gravitation and acceleration fields, ..."
End quote.
(The situation where the fields are equivalent
in acceleration we call them "orbits".)

Orbit is not necessary. Take a given test mass and
place it in a gravitational field and measure the
resulting acceleration. Take the same test mass and
take it far away from sources of gravity and apply
a force that produces the same acceleration as
before. The resulting effects will be
indistinguishable.

Yes, you're perfectly right once again.
Nevertheless, now your are talking about of the
second part of the problem (twice above).
The first part we've discussed so far was about circular
motion, orbits, gravity and centrifugal force.
The second part, you've just introduced now, it
is about radial motion towards the source. It is
the orthogonal problem, so to speak.
You know..., this is all about motions!
*Motion* is the most important matter of all.
Once you change the motion you change it all.

"Ken S. Tucker" escreveu na mensagem
...

I've a secret for you, gravity and centrifugal force
are not equivalent in nature. Gravity follows an
inverse SQUARED radius function. Centrifugal
follows an inverse LINEAR radius. Only for
stable orbits, where a body can be considered
to be in free-fall (no forces), one can make
such approximation.

As a little brat I've done the experiment
personally, I suspended a bucket and felt
the force. Then I removed some stones from
the bucket and twirled it.
I exchanged g-force for c-force+g-force.
Just do the experiment and get the feel.

Yep, you can bounce a rock around (not rock'n
roll) with a string and your personal experience
will give you exactly the same result. You're right.

The part you've missed was that centrifugal force
only exists if there is an infinitely rigid fixed point
in space where to apply the force.
Once you don't have such "magic" point, what you
have is a flying rocket.
The funny part is that planets look like rockets,
but behave like if there is a central force caused
by a string that keeps the orbit.
Such string is called gravity.

Double the lenght of the rope and you got the
double of the centrifugal force.
Double the orbit radius and you got 1/4 of the
gravity pull.
Do you see the difference now?

"El Enrrabadore-mor" escreveu na mensagem
...

"Ken S. Tucker" escreveu na mensagem
...

I've a secret for you, gravity and centrifugal force
are not equivalent in nature. Gravity follows an
inverse SQUARED radius function. Centrifugal
follows an inverse LINEAR radius. Only for
stable orbits, where a body can be considered
to be in free-fall (no forces), one can make
such approximation.

As a little brat I've done the experiment
personally, I suspended a bucket and felt
the force. Then I removed some stones from
the bucket and twirled it.
I exchanged g-force for c-force+g-force.
Just do the experiment and get the feel.

Yep, you can bounce a rock around (not rock'n
roll) with a string and your personal experience
will give you exactly the same result. You're right.

The part you've missed was that centrifugal force
only exists if there is an infinitely rigid fixed point
in space where to apply the force.
Once you don't have such "magic" point, what you
have is a flying rocket.
The funny part is that planets look like rockets,
but behave like if there is a central force caused
by a string that keeps the orbit.
Such string is called gravity.

Double the lenght of the rope and you got the
double of the centrifugal force.
Double the orbit radius and you got 1/4 of the
gravity pull.
Do you see the difference now?

Opss, forget to say that you must keep the same
angular frequency or speed. Remember this is a
thread about the speed of light in circular motion.

On May 4, 2:07 pm, "El Enrrabadore-mor"
wrote:
"Greg Neill" escreveu na sting.com..."El Enrrabadore-mor" wrote in message



Can you see how much out-of-topic you are?

Hey, *you're* the one who posted the nonsense,
"A system in resonance is a closed system that
exchanges no energy with surroundings. Energy
is conserved in resonance." I just pointed out
your misconceptions and errors.

Obviously you must agree that for a case of
electromagnetic radiation I'm right.
Not really. There is no such thing as resonance in a system
comprised of electromagnetic radiation only. By being in resonance we
are referring to the frequency of the elecromagnetic radiation
matching a resonance frequency in a bound system comprised of electric
charges. There is some type of displacement associated with the
electric charges.
No, we you solve the equation for displacement of the bound
system being acted upon by the electromagnetic radiation, the solution
has at least two components: a transient component and an
"equilibrium" component. In the theory of differential equations, the
transient component is called the homogenous solution and the
equilibrium component is called the inhomogenous component.
The exact strength and phase of the transient component vary on
the initial conditions on the displacement. The equilibrium solution
does not vary with initial conditions. Generally, if the displacement
starts at zero, the transient part dies away after a short time. Hence
it is often called the transient solution. The equilibrium component
never dies away.
I believe that you are referring to the equilibrium solution
when you say that "the system exchanges no energy with its
surroundings." The system usually refers to the bound system of
charges, which has a Q-factor. The transient part of course does
exchange energy with its surroundings. Otherwise it would never die
away. However, the equilibrium part is constant.
I think what you are ignoring is that the transient solution and
the equilibrium solution can cancel each other out when t=0. What
appears to be a build up of energy can also be described as the
transient part dying away. It took a second or two for that beer
bottle of yours to accumulate enough heat to melt, right? It also took
a few seconds for the energy of the microwaves to build up from the
oscillator in the microwave oven. You can't say that a system in
resonance doesn't exchange energy with its environment. In fact, your
example would seem to indicate the opposite of your conclusions. Your
microwave was plugged in, right? The socket supplied the power.
The bound system is partly characterized by what is called the Q-
factor. As Q increases, the time it takes for the transient component
to die increases. As Q increases, the amplitude of the equilibrium
part increases.
The time it takes takes for the transient solution die away
increases with the Q factor, The amplitude of the equilibrium solution
increases.
Some of your comments lead me to believe that you are trying to
develop a perpetual motion machine. If I misunderstood, then I
apologize. If I do understand you, then I have criticize your
presentation. Melting beer bottles in microwaves does not demonstrate
the creation of energy, or even the destruction of entropy. It
demonstrates the exchange of energy between an oscillator in the
microwave, the electromagnetic field in the microwave, and the
electric charges in your beer bottle. I recommend some other analogy.
I don't mean burning ants with the sun and a magnifying lens. Although
similar in some of the physics, it doesn't create energy either.